Optimal 3D Angular Resolution for Low-Degree Graphs

Transcription

1 Journal of Graph Algorithms and Applications vol. 17, no. 3, pp (2013) DOI: /jgaa Optimal 3D Angular Resolution for Low-Degree Graphs David Eppstein 1 Maarten Löffler 2 Elena Mumford 3 Martin Nöllenburg 4 1 Computer Science Department, University of California, Irvine, USA 2 Department of Information and Computing Sciences, Universiteit Utrecht, the Netherlands 3 Eindhoven, the Netherlands 4 Institute of Theoretical Informatics, Karlsruhe Institute of Technology, Germany Abstract We show that every graph of maximum degree three can be drawn without crossings in three dimensions with at most two bends per edge, and with 120 angles between all pairs of edge segments that meet at a vertex or a bend. We show that every graph of maximum degree four can be drawn in three dimensions with at most three bends per edge, and with angles, i. e., the angular resolution of the diamond lattice, between all pairs of edge segments that meet at a vertex or a bend. The angles in these drawings are the best possible given the degrees of the vertices. Submitted: July 2011 Reviewed: May 2012 Article type: Regular paper Revised: January 2013 Published: March 2013 Accepted: February 2013 Communicated by: H. Meijer Final: February 2013 This research was supported in part by the National Science Foundation under grants and , by the Office of Naval Research under MURI grant N , and by the German Research Foundation (DFG) under grant NO 899/1-1. addresses: eppstein@ics.uci.edu (David Eppstein) m.loffler@uu.nl (Maarten Löffler) mumford.elena@gmail.com (Elena Mumford) noellenburg@kit.edu (Martin Nöllenburg)

2 174 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs 1 Introduction Much past research in graph drawing has shown the importance of avoiding sharp angles at vertices, bends, and crossings of a drawing, as they make the edges difficult to follow [20]. There has been much interest in finding drawings where the angles at these features are restricted, either by requiring all angles to be at most 90 (as in orthogonal drawings [14] and right-angle crossing (RAC) drawings [1, 9, 10]) or more generally by attempting to optimize the angular resolution of a drawing, the minimum angle that can be found within the drawing [2, 5, 17 19, 21]. Three-dimensional graph drawing [11] opens new frontiers for angular resolution in two ways. First, in three-dimensional graph drawing, there is no need for crossings, as all graphs can be drawn without crossings; however, finding a compact layout that uses few bends and avoids crossings can sometimes be challenging. Second, and more importantly, in 3d there is a much greater variety in the set of ways that a collection of edges can meet at a vertex to achieve good angular resolution, and the angular resolution that may be obtained in 3d is often better than that for a two-dimensional drawing. For instance, in 3d, six edges may meet at a vertex forming pairwise angles of at least 90, whereas in 2d the same six edges would have an angular resolution of 60 at best. Several previous results in 3d graph drawing studied straight-line grid drawings with small volume but did not consider angular resolution [7, 16]. The problem of optimizing the angular resolution of a collection of edges incident to a single vertex in 3d is equivalent to the well-known Tammes problem of placing points on a sphere to maximize their minimum separation; this problem is named after botanist P. M. L. Tammes who studied it in the context of pores on grains of pollen [23], and much is known about it [6]. For graphs of degree five or six, the optimal angular resolution of a three-dimensional drawing is 90, as above, achieved by placing vertices on a grid and drawing all edges as grid-aligned polylines. The simplicity of this case has freed researchers to look for three-dimensional orthogonal drawings that, as well as optimizing the angular resolution, also optimize secondary criteria such as the number of bends per edge, the volume of the drawing, or combinations of both [4, 11, 13, 24]. Thus, in this case, it is known that the graph may be drawn with at most three bends per edge in an O(n) O(n) O(n) grid and with O(1) bends per edge in an O( n) O( n) O( n) grid [13]. For graphs of maximum degree five a tighter bound of two bends per edge is also known [24]; a well known open problem asks whether the same two-bend-per-edge bound may be achieved for degree six graphs [8]. No previous bounds are known on the volume and the number of bends of 3d drawings with angular resolution larger than 90, which obviously is possible only for graphs of maximum degree less than five. In two dimensions, every non-planar drawing has angular resolution at most 90, since the smallest angle at every crossing is at most 90. As stated before, in 3d, graphs of degree three and four may have angular resolution even better than 90 regardless of planarity, since every graph can be drawn without crossings in 3d. In particular, in the diamond lattice, a subset of the integer grid, the

3 JGAA, 17(3) (2013) 175 Figure 1: Left: The three-dimensional diamond lattice, from [15]. Right: A space-filling 3-regular graph with 120 angular resolution. edges are parallel to the long diagonals of the grid cubes and meet at angles of arccos( 1/3) 109.5, the optimal angular resolution for degree-four graphs (Figure 1, left). For graphs with maximum degree three, the best possible angular resolution at all degree-3 vertices is clearly 120 ; three edges with these angles are coplanar, but the planes of the edges at adjacent vertices may differ: for instance, Figure 1(right) shows an infinite space-filling graph in which all vertices are on integer grid points, all edges form face diagonals of the integer grid, and all vertices have 120 angular resolution. The primary questions we study in this paper are how to achieve optimal 120 angular resolution for 3d drawings of arbitrary graphs with maximum degree three, and optimal angular resolution for 3d drawings of arbitrary graphs with maximum degree four. We define angular resolution to be the minimum angle at any bend or vertex, matching the orthogonal drawing case, and we do not allow edges to cross. These questions are not difficult to solve without further restrictions (just place the vertices arbitrarily and use polylines with many bends to connect the endpoints of each edge) so we further investigate drawings that minimize the number of bends, align the vertices and edges of the drawing with the integer grid similarly to the alignment of the space filling patterns in Figure 1, and use a small total volume. 1.1 Results We consider two varieties of drawings: the grid-aligned case (where all vertices must be taken from a regular grid, depending on the degree, and all edges must be aligned with directions from the grid), and the free-form case (where vertices can be arbitrary points in space, and edges can go in any direction, as long as the angles are bounded from below by the required bound, depending on the degree). We show: All graphs of maximum degree four can be drawn in 3d with optimal angular resolution with at most three bends per edge, with all vertices

4 176 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs placed on an O(n) O(n) O(n) grid and with all edges parallel to the long diagonals of the grid cubes, i.e., using at most four different slopes. Every graph of maximum degree three can be drawn in 3d with optimal 120 angular resolution with at most two bends per edge. However, our technique for achieving this small number of bends does not use a grid placement and does not achieve good volume bounds. Every graph of maximum degree three has a drawing in 3d with 120 angular resolution, integer vertex coordinates, edges parallel to the face diagonals of the integer grid, at most three bends per edge, and polynomial volume. The edges use at most five different slopes. We believe that, as in the orthogonal case, it should be possible to achieve tighter bounds on the volume of the drawing at the expense of greater numbers of bends per edge. In the grid-aligned case, our results are optimal for degree-4 graphs, but it remains open whether all degree-3 graphs can be drawn with only two bends per edge. In the free-form case, our results are optimal for degree-3 graphs, but it remains open whether all degree-4 graphs can be drawn with only two bends per edge. Tables 1 and 2 summarize the state of the art and the remaining gaps. d = 1 d = 2 d = 3 0 bends YES (matching) NO ( ) NO ( ) 1 bend YES ( ) NO ( ) NO (Section 2.1) 2 bends YES ( ) NO ( ) 3 bends YES ( ) NO ( ) YES (Section 5) 4 bends YES ( ) NO (cycles) YES ( ) d = 4 d = 5 d = 6 0 bends NO ( ) NO (trivial) NO ( ) 1 bend NO ( ) NO (u-turn) 2 bends NO (Section 2.2) YES [24] 3 bends YES (Section 3) YES ( ) YES [13] 4 bends YES ( ) YES ( ) YES ( ) Table 1: Whether all graphs of degree d can be drawn aligned to a grid with angular resolution matching the optimal value needed for a single degree-d vertex, and with at most the given number of bends per edge. 2 Lower bounds In this section, we will observe some lower bounds on the number of bends that are necessary to draw graphs of a given degree. Let d be the maximum degree of a graph G and b the maximum number of bends per edge. For example, note that a graph of degree six can never be drawn on a grid with optimal angular

5 JGAA, 17(3) (2013) 177 d = 1 d = 2 d = 3 0 bends YES (matching) NO ( ) NO ( ) 1 bend YES ( ) NO ( ) NO (Section 2.1) 2 bends YES ( ) NO ( ) YES (Section 4) 3 bends YES ( ) NO ( ) YES ( ) 4 bends YES ( ) NO (cycles) YES ( ) d = 4 d = 5 d = 6 0 bends NO ( ) NO (trivial) NO (trivial) 1 bend NO (Section 2.2) 2 bends YES [24] 3 bends YES (Section 3) YES ( ) YES [13] 4 bends YES ( ) YES ( ) YES ( ) Table 2: Whether all graphs of degree d can be drawn with the optimal angular resolution for a single degree-d vertex, without regard to grid alignment, and with at most the given number of bends per edge. resolution and fewer than two bends per edges: consider the vertex v furthest in some Cartesian direction e. Vertex v has six outgoing edges, one of which leaves in direction e. However, this edge needs to connect to some other vertex, and it needs to turn by 90 at least twice to move back past v. On the other hand, this argument does not work in the free-form case: there it might happen that v has no edge leaving in direction e, and one bend per edge might suffice. The argument also does not extend to the case d = 5 on a grid: even though the optimal angle is still 90 in this case, each vertex has one direction in which no edge leaves. 2.1 Drawing K 4 We show that a drawing with optimal angular resolution does not always exist in the case d = 3 and b = 1, independent of whether the drawing is restricted to a grid or not. Consider K 4 ; all its vertices have degree three so they each need to be drawn with three coplanar stubs (possibly in a different plane for each vertex) leaving the vertex at 120 angles to each other. Now, note that every cycle in any graph drawing needs to have a total turning angle of at least 360. In the case d = 3, the turning angle at all vertices or bends is at most = 60. Therefore, every cycle needs at least six turning points. Now suppose K 4 has a drawing with one bend per edge. K 4 has four cycles of length three, each of which needs six turns, and a single cycle of this type can only be drawn in this way by using one bend on each of the three edges of the cycle. However, in this unique drawing of a 3 cycle, the six outgoing stubs from the three vertices are all coplanar, so all edges of the cycle need to be coplanar with the three vertices of the cycle. Since this is true for every triple of vertices, the whole drawing must lie in a single plane. Clearly, this is not possible with only one bend per edge.

6 178 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs z y x (1, 1, 0) (4, 1, 3) (0, 0, 0) ( 2, 0, 2) (3, 1, 6) (4, 1, 5) (0, 1, 1) (5, 2, 5) ( 2, 3, 5) (3, 4, 1) ( 1, 4, 5) (0, 1, 6) (5, 5, 2) y (5, 5, 0) ( 1, 6, 3) (2, 6, 0) x Figure 2: A two-bend drawing of K 4 with 120 angular resolution (left) and its two-dimensional projection (right). This argument shows that not every graph of degree three can be drawn with one bend per edge, both in the grid-aligned and in the free-form setting. For the free-form setting, this bound is tight, as we show in Section 4. In the grid-aligned case, it remains an open problem whether every graph can be drawn with two bends per edge. The complete graph K 4 does admit such a drawing, as we show in Figure Drawing K 5 We will now consider grid-aligned drawings of K 5. Note that the grid in this case consists of lines with four orientations: the four main diagonals of a cube. This leads to eight possible directions for a piece of an edge, which we will identify with the eight corners of a cube. Whenever an edge makes a turn (whether at a vertex or a bend), its direction changes from one cube corner to an adjacent cube corner, which corresponds to a turning angle of On a grid, vertices of a degree-4 graph can have two types: type-1 vertices have stubs in directions (1, 1, 1), (1, 1, 1), ( 1, 1, 1) and ( 1, 1, 1); while type-2 vertices have stubs in the opposite directions ( 1, 1, 1), ( 1, 1, 1), (1, 1, 1) and (1, 1, 1). In this section, we will call directions (1, 1, 1) and ( 1, 1, 1) blue, directions (1, 1, 1) and ( 1, 1, 1) purple, directions ( 1, 1, 1) and (1, 1, 1) red, and directions ( 1, 1, 1) and (1, 1, 1) orange. Now, notice that an edge with no bends connects a type-1 vertex with a type-2 vertex, an edge with one bend connects same-type vertices, an edge with two bends connects different-type vertices again, etc. The graph K 5 has five vertices. By the pigeon hole principle, there must be three of them that have the same type. Assume there is a grid-aligned drawing of K 5 with at most two bends per edge. Then the edges that connect vertices of the same type can have only one bend. This means that the cycle of same-type vertices must be drawn as a hexagon H, as shown in Figure 3. Note that H does not lie in a plane.

7 JGAA, 17(3) (2013) 179 w z u x y v Figure 3: A grid-aligned cycle of length three, which is necessarily contained in all drawings of K 5 that are aligned to the grid and have optimal angular resolution. The colored lines are the allowed directions; the cubes are shown for reference. Now, consider the three vertices u, v, and w of H. Assume the drawing is translated and rotated such that u lies on the x-axis, v lies on the y-axis, and w lies on the z-axis, as in Figure 3. Each vertex has two remaining stubs, one pointing in the orange direction (1, 1, 1), and one pointing in a unique direction. Again by the pigeon hole principle, at least two of the three unique stubs (blue, red, and purple) must connect to the same vertex s. Since the drawing is symmetric (up to edge piece lengths), we can assume without loss of generality that the red stub of u and the purple stub of v both connect to s. Claim 1 With the assumptions given above, s lies in the halfspace z < 0. Furthermore, the edge from u reaches s in the purple direction, and the edge from v reaches s in the red direction. Proof: Consider the plane Γ bo : x = y (the plane containing the origin and spanned by the blue and orange directions). Note that u and v lie on opposite sides of Γ bo. Suppose s lies on the side of v (the other case is symmetric). Then the edge connected to the red stub of u, which points in the direction (1, 1, 1), has to turn enough to cross Γ bo, which is only possible with two bends if it turns to the purple direction ( 1, 1, 1). To reach the purple direction it has to go via

8 180 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs the orange direction ( 1, 1, 1) or the blue direction (1, 1, 1); in either case the edge never travels in the positive z-direction, so since u has z-coordinate 0, s must lie in the halfspace z < 0. Finally, we need to argue that the edge that connects to the stub from v reaches s in the red direction (1, 1, 1). Suppose the edge from u uses the orange direction (again, the other case is symmetric). Consider the plane Γ op : x = z (spanned by orange and purple), and note that s lies on the same side of Γ op as u (since the edge from u never crosses it), but v lies on Γ op. So, to reach s, the edge from v needs to leave Γ op towards the side of u. If it does not do so by turning all the way to the red direction (1, 1, 1) (which is what we aim to show), then it must use the blue direction (1, 1, 1). Now, consider the plane Γ bp : y = z (spanned by blue and purple). Since u lies on Γ bp, and the edge from u never crosses it, the edge from v needs to cross it. But, since it first turned from the purple direction ( 1, 1, 1) to the blue direction (1, 1, 1), both of which span the plane Γ bp, the only way to cross Γ bp is again to turn to the red direction (1, 1, 1). However, now s also needs to connect to one of the stubs of w. The remaining stubs of s leave in the blue direction (1, 1, 1) and the orange direction ( 1, 1, 1), while the remaining stubs of w leave in the blue direction ( 1, 1, 1) and the orange direction (1, 1, 1). Since w has a positive z-coordinate, and s has a negative z-coordinate, the only way to get from w to z with only two bends is if the middle piece of the edge travels in a direction with negative z-coordinate: either blue-orange-blue or orange-blue-orange. Suppose it uses the pattern orange-blue-orange (the other case is symmetric). Now, consider once more the plane Γ bp : as we have seen, s must lie below this plane. However w lies above Γ bp, and an edge traveling from w in only the orange direction (1, 1, 1) and the blue direction (1, 1, 1) cannot cross it. We conclude that a drawing of K 5 with only two bends per edge does not exist. This argument shows that not every graph of degree four can be drawn with two bends per edge in the grid-aligned setting, and we show in Section 3 that this bound is tight. In the free-form setting, there is a drawing of K 5 with two bends per edge, as we show in Figure 4. It remains an open problem whether such a drawing exists for every graph of degree four. 3 Three-bend drawings of degree-four graphs on a grid Our technique for three-dimensional drawings of degree-four graphs with angular resolution and three bends per edge is based on lifting certain twodimensional drawings of the same graphs, with angular resolution 90 and two bends per edge. The three-dimensional vertex placements are all on the plane z = 0, essentially unchanged from their two-dimensional placements, but the edges are raised and lowered above and below the plane to avoid crossings and improve the angular resolution.

9 JGAA, 17(3) (2013) 181 v 5 v 3 z y x v 1 v 1 : (0, 0, 0) v 2 : (1.87, 1.59, 0) v 3 : (1.87, 1.59, 0) v 4 : (0.75, 0, 0.82) v 5 : (0.75, 0, 0.82) v 4 v 2 Figure 4: A two-bend drawing of K 5 with angular resolution, drawn in the free-form setting. 3.1 Two-dimensional drawings Our two-dimensional orthogonal drawing technique uses ideas from previous work on drawing degree-four graphs with bounded geometric thickness [12]. We begin by augmenting the graph with dummy edges and a constant number of dummy vertices if necessary to make it a simple 4-regular graph, find an Euler tour in the augmented graph, and color the edges alternately red and green in their order along this path. In this way, the red edges and the green edges each form 2-regular subgraphs [22] consisting of disjoint unions of cycles. We denote the number of red (green) cycles by m red (m green ). Next, we draw the red subgraph so that every red cycle passes horizontally through its vertices with two bends per edge, and we draw the green subgraph so that every green cycle passes vertically through its vertices with two bends per edge. We can do that by using the cycle ordering within each of these two subgraphs as one of the two Cartesian coordinates for each point. More precisely, we do the following. We define the green order of the vertices of the graph to be an order of the vertices such that the vertices of each green path or cycle are consecutive; we define the red order the same way. Let r green (v) 0 be the rank of a vertex v in some green order, and r red (v) be its rank in some red order. We further order the red and green cycles and define c red (v) 0 and c green (v) 0 to be the ranks in the two cycle orders of the red and green cycles to which v belongs. We embed the vertices on a (2n + 2m green 4) (2n + 2m red 4) grid such that the x-coordinate of each vertex is 2r green (v) + 2c green (v), and its y-coordinate is 2r red (v) + 2c red (v). Let v 1,...v k be the vertices of a green cycle C in the green order. We embed C as follows. We mark each end of each edge with a plus or a minus such that at

10 182 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 v 2 v 2 v 5 v 5 v 9 v 9 v 7 v 7 v 1 v 1 v 4 v 4 v 8 v 8 v 3 v 3 v 6 v 6 v 10 v 10 Figure 5: A 4-regular graph with 10 vertices embedded according to the decomposition into disjoint red and green cycles. every vertex exactly one end is marked with a plus and exactly one with a minus. We then would like to embed C in such a way that plus would correspond to the edge entering the vertex from above and a minus corresponds to the edge entering the vertex from below. Note that every edge whose two ends are marked the same can be embedded in this way with two bends. Whenever the marks alternate along the edge one can only embed it with two bends if the lower end (the end incident to a vertex with smaller y-coordinate) is marked with plus. We next describe how to label C so that it has a 2-bends-per-edge embedding respecting the labeling. If k is even, we mark both ends of the edge (v 1, v 2 ) with pluses. If k is odd, we mark the higher end of (v 1, v 2 ) with a minus and its lower end with a plus. In both cases there is a unique way to label the rest of the edges such that both ends of each edge have the same signs and the labels alternate at every vertex. To complete our 2d embedding we draw all edges consistently with the labeling as follows. Each edge (v i, v i+1 ) is placed such that the y-distance of its horizontal segment to one of the vertices is 1. If the last edge (v 1, v k ) is labeled negatively, its horizontal segment is drawn on the grid line one unit below the lowest vertex or bend of C. Similarly, if (v 1, v k ) is labeled positively, the horizontal segment is drawn one unit above the highest part of C. See Figure 5 for an illustration. Lemma 1 There exist red and green orders that have the following properties:

11 JGAA, 17(3) (2013) 183 no two edges of the same color intersect; a vertex lies on an edge if and only if it is incident to the edge; no midpoint of an edge coincides with a bend of the edge; the embedding fits on a (2n + 2m green ) (2n + 2m red ) grid. Proof: Green edges connecting consecutive vertices in the green order of the same cycle C are trivially disjoint. The horizontal segment of the edge connecting the first and the last vertex of C is placed below or above all other edges of C. Two different green components are disjoint because the edges of every component are contained inside the vertical strip defined by its first and last vertices and components are ordered along the x-axis. The argument for red edges is symmetric. Since all the vertices have distinct x-coordinates, and every green vertical segment has a vertex at one of its ends we can conclude that every vertex is incident to at most two vertical green segments. Every green horizontal segment has odd y-coordinate and every vertex has even y-coordinate hence a green horizontal segment cannot contain a vertex. The argument for red edges is symmetric. For arbitrary red and green vertex orders it is possible that the midpoint of an edge coincides with one of its bends. We show that there are red and green vertex orders for which this is not the case. For every edge whose ends are labeled differently we can always place the horizontal segment such that the midpoint of the edge does not coincide with a bend. For edges whose ends have the same label it is easy to see that the midpoint coincides with a bend if and only if the vertical distance and the horizontal distance of its vertices are equal. Apart from the last edge in each green cycle the horizontal distance between every two adjacent vertices v i and v i+1 is 2. We claim that the vertical distance between v i and v i+1 is larger than 2 since otherwise v i and v i+1 are adjacent in a red cycle which contradicts the assumption that the 4-regular graph is simple. Note that this is the reason why different components are spaced by at least 4 units. Finally consider the last edge (v 1, v k ) of a cycle C with vertices v 1,..., v k. The horizontal distance of v 1 and v k is 2k 2. If their vertical distance equals 2k 2 as well, we cyclically shift the green order of the vertices in C by moving v k to the vertical grid line of v 1 and shifting each of v 1,..., v k 1 two units to the right. Now (v k, v k 1 ) is the last edge of C. We perform this shifting until the vertices of the last edge no longer have vertical distance 2k 2. Since every vertex has an exclusive y-coordinate there is at least one edge with this property in C. The local shifting of C does not influence other parts of the drawing. The argument for red cycles is analogous. The vertices lie on (2n + 2m green 4) (2n + 2m red 4) grid, and each grid line with coordinate 2k contains exactly one vertex. The lowest vertex is incident to a green edge with a horizontal segment at the height 1; the highest one is incident to a green edge with a horizontal segment at the height 2n + 2m red 3.

12 184 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs One of the green edges connecting the first and last vertices of some cycle can lie one grid line below the height 1 or one grid lines above 2n + 2m red Lifting to three dimensions It remains to lift the 2d drawing described above into three dimensions. We first rotate the drawing by 45 ; this expands the grid size to (4n + 4m green ) (4n + 4m red ). The vertices themselves stay in the plane z = 0, but we replace each edge by a path in 3d that goes below the plane for the red edges and above the plane for the green edges, eliminating all crossings between red and green edges. The path for a green edge goes upwards along the long diagonals of the diamond lattice cubes until its midpoint, where it has a bend and turns downwards again. The lifted images of the two bends in the underlying 2d edge remain bends in the 3d path and hence we get three bends per edge in total. The red edges are drawn analogously below the plane z = 0. Since in the original 2d drawing every edge has even length, the midpoint of every edge is a grid point and hence the lifted midpoint is also a grid point of the diamond lattice. By Lemma 1 a midpoint of an edge never coincides with a 2d bend and hence all bend angles as well as the vertex angles are diamond lattice angles. We note that by following the long diagonals in the diamond lattice our edges use only four different slopes. Finally, we remove all the edges we added to make the graph 4-regular. Considering the longest possible red and green edges the total grid size is at most (4n + 4m green ) (4n + 4m red ) (12n + 6m green + 6m red ). We note that m green, m red n/3 since every component is a cycle. This yields the following theorem. Theorem 1 Every graph G with maximum vertex degree four can be drawn in a 3d grid of size 16n/3 16n/3 16n with angular resolution 109.5, three bends per edge and no edge crossings. The drawing uses at most four different edge slopes. 4 Two-bend drawings of degree-three graphs The main idea of our algorithm for drawing degree-three graphs with optimal angular resolution and at most two bends per edge is to decompose the graph into a collection of vertex-disjoint cycles. Each cycle of length four or more can be drawn in such a way that the edges with exactly one endpoint in the cycle all attach to it via segments that are parallel to the z-axis (Lemma 4). By placing the cycles far enough apart in the z-direction, these segments can be connected to each other such that the resulting edge has at most two bends per edge. However, several issues complicate this method: Cycles of length three cannot be drawn in the same way, and must be handled differently (Lemma 2). Our method for eliminating cycles of length three does not apply to the graph K 4, for which we need a special-case drawing (Lemma 3).

13 JGAA, 17(3) (2013) 185 Figure 6: Y transformation of a graph G containing a triangle, and undoing the transformation to find a drawing of G (Lemma 2). Top: the contracted vertex has degree three, and is replaced by a hexagon. Bottom: the contracted vertex has degree two, and is replaced by a heptagon. Although Petersen s theorem [3, 22] can be used to decompose every bridgeless cubic graph into cycles and a matching, it is not suitable for our application because some of the matching edges may connect two vertices in a single cycle, a case that our method cannot handle. In addition, we wish to handle graphs that may contain bridges. Therefore, we need to devise a different decomposition algorithm. However, with our decomposition, the graph obtained by removing all cycle edges is a forest rather than just a matching, and again we need additional analysis to handle this case. Lemma 2 Let G be a graph with maximum degree three containing a triangle uvw. If uvw is not part of any other triangle, let G be the result of contracting uvw into a single vertex (that is, performing a Y transformation on G). Otherwise, if there is a triangle vwx, let G be the result of contracting uvwx into a single vertex. If G can be drawn in 3d with two bends per edge and with angles of at least 120 between the edges at each vertex or bend, then so can G. Proof: First we consider the case that G is obtained by collapsing uvw. The edges incident to the merged vertex uvw must lie in a plane in any drawing of G. If uvw has degree zero, one, or three in G, or if it has degree two and is drawn with angular resolution exactly 120, then we may draw G by replacing uvw by a small regular hexagon in the same plane, with at most one bend for each of the three triangle edges (Figure 6, top). If the merged vertex uvw has degree two in G and is drawn with angular resolution greater than 120, we may replace it by a small heptagon (Figure 6, bottom). The case that G is obtained by collapsing four vertices uvwx is similar: the collapsed vertex may be replaced by a pair of regular hexagons or irregular heptagons, meeting edge-to-edge. The four vertices uvwx are placed at the points where these two polygons meet the other edges of the drawing and the

14 186 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs Figure 7: The embedding of a cycle with degree-one neighboring vertices described by Lemma 4. Left: the xy-projection of the cycle; cycle vertices are indicated as large hollow circles and bends are indicated as small black disks. Right (at a larger scale): the xz-projection of the portions of the embedding corresponding to the two horizontal bottom sides of the xy-projected polygon. two endpoints of the edge where they meet each other; the edge vw has no bends and the other edges all have one or two bends. Lemma 3 The graph K 4 may be drawn in 3d with all vertices on integer grid points, angular resolution 120, and at most two bends per edge. Proof: See Figure 2. Lemma 4 Let G be a graph with maximum degree three, consisting of a cycle C of n 4 vertices together with some number of degree-one vertices that are adjacent to some of the vertices in C. Suppose also that each degree-one vertex in G is labeled with the number +1 or 1. Then, there is a drawing of G with the following properties: All vertices and bends have angular resolution at least 120. All edges of C have at most two bends. All edges attaching the degree-one vertices to C have no bends. Every degree-one vertex has the same x and y coordinates as its (unique) neighbor, and its z coordinate differs from its neighbor s z coordinate by its label. Thus, all edges connecting degree-one vertices to C are parallel to the z-axis, all positively labeled vertices are above (in the positive z-direction from) their neighbors, and all negatively labeled vertices are below their neighbors. No three vertices of C project to collinear points in the (x, y)-plane. Proof: As shown in Figure 7, we draw C in such a way that it projects onto a polygon P in the xy-plane, with 135 angles and with sides parallel to the coordinate axes and at 45 angles to the axes. There are polygons of this type

15 JGAA, 17(3) (2013) 187 with a number of sides that can be any even number greater than seven; we choose the number of sides of P so that at least one and at most two vertices of C can be assigned to each axis-parallel side of the polygon. (E. g., when C has from four to eight vertices, P can have eight sides, but when C has more vertices P must be more complex.) We assign the vertices of C consecutively to the axis-parallel sides of P, in such a way that at least one vertex of C and at most two vertices are assigned to each axis-parallel side. If one vertex is assigned to a side, it is placed at the midpoint of that side, and if two vertices are assigned to a side of length l, then they are placed at distances of l/4 from one endpoint of the side, as measured in the xy plane, with a bend at the midpoint of the side. In three dimensions, the diagonal sides of P are placed in the plane z = 0. For every axis-parallel side of P of length l containing k vertices of C, we place the vertices with no degree-one neighbor or with a positively labeled neighbor at elevation z = l/(2k 3), and the vertices with a negatively labeled neighbor at elevation z = l/(2k 3), so that the portion of C that projects onto a single side of P forms a polygonal curve with angles of exactly 120. The degree-one neighbors of the vertices in C are then placed above or below them according to their signs. With this embedding, each vertex of C gets angular resolution exactly 120. Every two consecutive vertices of C that are assigned to the same side of P are separated either by zero bends (if their neighbors have opposite signs) or a single bend (if their neighbors have the same signs). Two consecutive vertices of C that belong to two different sides of P are separated by two bends at two of the corners of P ; these bends have angles of arccos( 3/8) By adjusting the lengths of the sides of P appropriately, we may ensure that no three vertices of C project to collinear points in the xy-plane. The main idea of our drawing algorithm is to use Lemma 4, and some simpler cases for individual vertices, to repeatedly extend partial drawings of the given graph G until the entire graph is drawn. We define a vertically extensible partial drawing of a set S of vertices of G to be a drawing of the subgraph G[S] induced in G by S, with the following properties: The drawing of G[S] has angular resolution 120 or greater and has at most two bends per edge. Each vertex in S has at most one neighbor in G \ S. If a vertex v in S has a neighbor w in G \ S, then w could be placed anywhere along a ray in the positive z-direction from v, producing a drawing of G[S {w}] that remains non-crossing, continues to have angular resolution 120 or greater, and has no bends on edge vw. We call the ray from v the extension ray for edge vw. No three extension rays are coplanar. For instance, if C is a chordless cycle of length four or greater in G, then by Lemma 4 there exists a vertically extensible partial drawing of C. More, the

16 188 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs Figure 8: Extending a vertically extensible drawing by adding a cycle. same lemma may be used to add another cycle to an existing vertically extensible partial drawing (Figure 8): Lemma 5 For every vertically extensible drawing of a set S of vertices in a graph G of maximum degree three, and every chordless cycle C of length four or more in G \ S, there exists a vertically extensible drawing of S C. Proof: For each vertex v in C that has a neighbor w in G, replace w with a degree-one vertex that has label 1 if w S and +1 if w / S. Apply Lemma 4 to find a drawing of C that can be connected in the negative z-direction to the neighbors of C in S, and in the positive z-direction for the remaining neighbors of C. Translate this drawing of C in the xy-plane so that, among the extension rays of S and the vertices of C, there are no three points and rays whose projections into the xy-plane are collinear and so that, when projected onto the xy-plane, the extension rays of S (points in the xy-plane) are disjoint from the projection of the drawing of C. For each extension ray of S that connects a vertex v of S to a vertex w in C, draw a two-bend path with 120 bends in the plane containing the extension ray and w, such that the final segment of the path has the same x and y coordinates of w. By making the transverse section of this path be far enough away from S in the positive z-direction, it will not intersect any other features of the existing drawing, and it cannot cross any of the other extension rays due to the requirement that no three of these rays be coplanar. If C is translated in the positive z-direction farther than all of the bends in these paths, it can be connected to S to form a vertically extensible drawing of S C, as required. Lemma 6 For every vertically extensible drawing of a set S of vertices in a graph G of maximum degree three, and every vertex v in G \ S with at most two neighbors in S and at most one neighbor in G \ S, there exists a vertically extensible drawing of S {v}. Proof: If v has no neighbors in S, then v may be placed anywhere on any z-parallel line that does not pass through a feature of the existing drawing and is

17 JGAA, 17(3) (2013) 189 v w u w t v u Figure 9: Left: Adding a vertex v with two neighbors u and w in S and one neighbor in G \ S to a vertically extensible drawing (shown in the plane of the extension rays of u and w). Right: Adding a vertex v with three neighbors t, u, and w in S (shown in the xy-plane). The three segments incident to v are parallel to the xy plane and the three remaining transverse segments form 120 angles to the extension rays of t, u, and w. The bends where these transverse segments meet their extension rays are shown on top of the three points t, u, and w. not coplanar with any two existing extension rays. If v has a single neighbor w in S, then v may be placed anywhere on the extension ray of wv. In the remaining case, v connects to two extension rays of S. Within the plane of these two rays, we may connect v to these two rays by transverse segments at 120 angles to the rays. By placing v far enough in the positive z-direction, these transverse segments can be made to avoid all existing features of the drawing. The extension ray from v can lie on any line parallel to and between the lines of the two incoming extension rays; only finitely many of these lines lead to co-planarities with other extension rays, so it is always possible to place v avoiding any such coplanarity. As shown in Figure 9(left), this construction produces one bend on each edge into v. Lemma 7 For every vertically extensible drawing of a set S of vertices in a graph G of maximum degree three, and for every vertex v in G \ S that has three neighbors t, u, and w in S, there exists a vertically extensible drawing of S {v}. Proof: Suppose that tu is the longest edge of the triangle formed by the projections of t, u, and w into the xy plane. Then, as a first approximation to the position of v in the xy-plane, let the (two-dimensional) point v be placed on edge tu of this triangle, at the point where v w is perpendicular to tu. We adjust this position along edge tu, keeping the angle between v w and tu close to 90 in order to ensure that line segment v w does not pass through the two-dimensional projection of any extension ray. Then, we replace v by three short line segments at 120 angles to each other meeting the three line segments v t, v u, and v w at angles of 150, 150, and close to 180. Let v be the point where these three short line segments meet. This configuration can be lifted into three-dimensional space by placing v and the three edges that attach to it in a plane perpendicular to the z-axis, and by replacing the remaining portions of line segments v t, v u, and v w by

18 190 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs transverse segments that make 120 angles with the extension rays of t, u, and w. There are two bends per edge: one at the point where the extension ray of t, u, or w meets a transverse segment, and one where a transverse segment meets one of the horizontal segments incident to v. The angles at the bends on the extension rays of t, u, and w are all exactly 120, and the angles at the other bends on the paths connecting t and u to w are arccos(3/4) As long as segment v w stays within 54 of perpendicular to tu in the xy-plane, the angle at the final remaining bend will be at least 120. The construction of Lemma 7 is illustrated in Figure 9(right). Theorem 2 Any graph G of maximum degree three has a 3d drawing with 120 angular resolution and at most two bends per edge. Proof: While G contains a triangle, apply Lemma 2 to simplify it, resulting in either K 4 or a triangle-free graph G. If this simplification process leads to K 4, draw it according to Lemma 3. Otherwise, starting from S =, we repeatedly grow a vertically extensible drawing of a subset S of G until all of G has been drawn. If G \ S contains a vertex with at most one neighbor in G \ S, then either Lemma 6 or Lemma 7 applies and we can add this vertex to the vertically extensible drawing. Otherwise, all vertices in G \ S have two or more neighbors in G \ S, so G \ S contains a cycle. Let C be the shortest cycle in G \ S; it has length at least four (because we eliminated all triangles) and no chords (because a chord would lead to a shorter cycle) so we may apply Lemma 5 to incorporate it into the vertically extensible drawing. Once we have included all vertices in the vertically extensible drawing, we have drawn all of G, and we may reverse the transformations performed according to Lemma 2 to produce a drawing of G. 5 Three-bend drawings of degree-three graphs on a grid The approach described in the previous section does not extend to the grid-aligned setting; in particular, there seems to be no obvious way to make Lemma 7 work in this case. Therefore, in this section we provide an algorithm for embedding a degree-three graph on a grid, using a similar approach to Theorem 2 but with up to three instead of two bends per edge. The grid will consist of the face diagonals of the cubes in a regular grid of cubes. First of all, we will make a change of coordinates that allows us an easier description. Define the xy-plane to be the plane spanned by the edges e X = (0, 1, 1) and e Y = (1, 1, 0) and the yz-plane to be the plane spanned by vectors e Y and e Z = (1, 0, 1). See Figure 10. We will draw the different parts of the drawing in either a horizontal plane (parallel to the xy-plane) or in a vertical plane (parallel to the yzplane). The edges we use in the xy-plane are parallel to an edge in the set E XY = {e X, e Y, e X e Y } and they all form angles of 120. Similarly, in the

19 JGAA, 17(3) (2013) 191 z e X(0, 1, 1) e Z(1, 0, 1) y e Y (1, 1, 0) x Figure 10: The three base vectors e X, e Y, e Z. yz-plane, all edges are parallel to an edge in E Y Z = {e Y, e Z, e Y e Z }. This means that only five different slopes are used. Furthermore, all edge segments have integer lengths. The construction works similarly to the one described in Section 4. In particular, we use exactly the same decomposition of the graph into cycles and trees, and we still draw every cycle in a different horizontal plane and extend the drawing in z-direction with every new cycle. However, there are some important differences. First of all, we no longer point the extension rays up (in the z-direction), but to the right (the y-direction), within the plane in which we draw the cycle. As a result, the drawing of a cycle is completely flat. Then, we draw the trees in vertical planes through the extension rays of the respective vertices. Figure 11(b) and (c) shows the general idea. (a) (b) x z y y Figure 11: (a) A cycle in a horizontal plane; (b) the tree structure connecting the extension rays in a set of three neighboring vertical planes (indicated by different colors). Lemma 8 Let G be a graph with maximum degree three, consisting of a cycle C of n 3 vertices together with some number of degree-one vertices that are adjacent to some of the vertices in C. Let x 1,..., x n be a set of distinct even integers bounded by O(n). Then, there is a drawing of G in the xy-plane with the following properties: All vertices and bends have angular resolution 120. All edges of C have at most three bends. All edges of G are parallel to an edge in E XY. For every degree-one vertex v = (x v, y v, z v ) and its neighbor u = (x u, y u, z u ) we have x v = x u, y v = y u + 1, and z v = z u.

20 192 Eppstein et al. Optimal 3D Angular Resolution for Low-Degree Graphs The drawing fits into a grid of size O(n) O(n 2 ). The x-coordinate of a vertex v i of C is x i, for all 1 i n. Proof: We embed each cycle in a similar manner as in Section 3. We label each end of each edge of the cycle with a plus or a minus sign such that at every vertex exactly one end is marked with a plus sign and exactly one with a minus sign. We construct this labeling in exactly the same way as in Section 3. Suppose that the y-axis of the xy-plane points horizontally to the right. We then would like to embed the cycle in the hexagonal grid of the xy-plane in such a way that edges labeled with a plus sign enter the vertex from above and edges labeled with a minus sign enter the vertex from below. Moreover, the edge-segment entering a vertex from below is e X -parallel and the edge-segment entering a vertex from above is (e X e Y )-parallel. With this orientation for the edges, every edge whose ends have identical labels can be embedded with exactly three bends. However, an edge that has opposite signs at its two ends can only be embedded with three bends if its lower end (the end incident to a vertex with smaller x-coordinate) is the end labeled with a plus sign. We then embed the edges and vertices as follows. We place the vertex v 1 at some point (x 1, y 1 ) in the xy-plane, where x 1 is its given x-coordinate. If the cycle contains an odd number of vertices then the first edge (v 1, v 2 ) is labeled with two opposite signs, and is drawn as follows assuming that v 1 is the lower of the two vertices. From v 1 we draw an e X e Y -parallel edge segment followed by an e X -parallel segment of equal length such that we reach the e Y -parallel line at x = x 2. We place v 2 at that position. If i > 1 or if i = 1 and the cycle has an even number of vertices, then edge v i, v i+1 is labeled with two plus signs or two minus signs. In the case that it is labeled with two plus signs, we start drawing an e X e Y -parallel edge segment from v i followed by an e X -parallel segment of equal length until we reach an e Y -parallel line that is two units above the higher of the two vertices. If these edge segments intersect any previous part of the drawing we may need to spread the drawing in the e Y -direction by a distance of O(n) that is added to the length of an e Y -parallel edge. From there we add a unit-length e Y -parallel segment and another e X e Y -parallel segment until we reach the e Y -parallel line with x = x i+1. This is where we place v i+1. We proceed symmetrically for any edge marked with two minuses. For the edge (v 1, v n ) the only difference is that its e Y -parallel segment is placed either two units below the lowest point or two units above the highest point of the drawing according to its labels. Finally, we embed each degree-one vertex one unit to the right of its cycleneighbor. See Figure 11(b) for an illustration. Note that the size of the grid for drawing G is linear in the x-direction but in the worst case quadratic in the y-direction.

21 JGAA, 17(3) (2013) 193 As in our two-bend non-grid embedding for degree-three graphs, our overall embedding algorithm begins by finding and embedding a chordless cycle of a given graph G and then extends partial drawings of our graph G using Lemma 8 until we obtain the drawing of the entire graph. We define an extensible partial grid-drawing of a set S of vertices of G to be a crossing-free grid-drawing of the subgraph G[S] induced in G by S, with the following properties: The drawing of G[S] has angular resolution 120. Each vertex in S has at most one neighbor in G \ S. Each vertex in G \ S has at most one neighbor in S. If a vertex v in S has a neighbor w in G \ S, then we can draw an edge (v, w) with at most three bends of 120 that starts with an e Y -parallel edge segment called the extension ray of v. The placement of (v, w) and w is such that the resulting drawing of G[S {w}] has angular resolution 120 and remains extensible and non-crossing. For every x-coordinate value x 0 there is at most one vertex v in the vertical plane through x 0 with an active extension ray, i. e., an extension ray that is not yet part of an actual edge (v, w) since w is still a vertex in G \ S. All vertices in S have even z-coordinates. One difference between these properties and the ones used for our two-bend drawings is the requirement that each vertex in G \ S have at most one neighbor in S. To meet this requirement, when we add a cycle to the drawing, we will also add more vertices until this requirement is met. To formalize this, define the double-adjacency closure of a set of vertices S in a graph G to be the smallest superset W (S) S such that every vertex in G \ W (S) that is adjacent to W (S) has two other neighbors in G \ W (S). The double-adjacency closure of S may be obtained by initializing a variable set W to be empty and then repeatedly adding to W any vertex in G \ (S W ) that has at least two neighbors in S W until no more such vertices exist; once this process converges, the double-adjacency closure W (S) is S W. Lemma 9 For every extensible grid-drawing of a set S of vertices in a graph G of maximum degree three, and every chordless cycle C in G \ S, there exists an extensible grid-drawing of W (S C), where W (S C) is the double-adjacency closure of S C. Proof: For each vertex v in C that has a neighbor w in G, replace w with a degreeone vertex. We next determine the x-coordinate x v for each vertex v of C as follows. If v is adjacent to some vertex w in S or there is a third vertex u in G \ (S C) that is adjacent to both v and some vertex w in S then we set x v to be the x w (we can do that because w is unique). The same applies if there is a vertex u in G \ (S C) that is adjacent to both v and an already placed vertex w